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Let \(l_1\) and \(l_2\) be two non-vertical lines.
The slope of line \(l_1\) is \(m_1\), and line \(l_2\) is \(m_2\).
Let the inclination of \(l_1\) be \(\theta_1\) and \(l_2\) be \(\theta_2\).
Then, \(m_1 = \tan \theta_1\) and \(m_2 = \tan \theta_2\).
Assume \(l_1\) and \(l_2\) are perpendicular lines.
In \(\Delta ABC\):
\(\angle A = \theta_1\) and \(\angle C = 90^\circ\)
Then, \(\angle B = 180^\circ - \angle A - \angle C\) [by Angle sum property].
\(\angle B = 180^\circ - \theta_1 - 90^\circ\)
\(\angle B =90^\circ - \theta_1\)
Measuring slope of \(l_2\) through angle \(\theta_2\) and \(90^\circ - \theta_1\) are opposite of each other.
\(\Rightarrow \tan \theta_2 = - \tan (90^\circ - \theta_1)\)
\(\Rightarrow \tan \theta_2 = - \frac{\sin(90^\circ - \theta_1)}{\cos(90^\circ - \theta_1)}\)
[Using the trigonometric identities \(\sin (90^\circ - A) = \cos A\) and \(\cos (90^\circ - A) = \sin A\)]
\(\Rightarrow \tan \theta_2 = - \frac{\cos \theta_1}{\sin \theta_1}\)
\(\Rightarrow \tan \theta_2 = - \cot \theta_1\)
\(\Rightarrow \tan \theta_2 = - \frac{1}{\tan \theta_1}\)
\(\Rightarrow \tan \theta_1 \cdot \tan \theta_2 = -1\)
\(\Rightarrow m_1 \cdot m_2 = -1\)
Conversely:
Let \(l_1\) and \(l_2\) be two non-vertical lines with slopes \(m_1\) and \(m_2\), respectively, such that \(m_1 \cdot m_2 = -1\).
\(\Rightarrow \tan \theta_1 \cdot \tan \theta_2 = -1\)
\(\Rightarrow \tan \theta_1 = - \frac{1}{\tan \theta_2}\)
\(\Rightarrow \tan \theta_1 = - \cot \theta_2\)
\(\Rightarrow \tan \theta_1 = - \tan (90^\circ - \theta_2)\) [Using \(\cot (90^\circ - A) = \tan A\)]
\(\Rightarrow \tan \theta_1 = \tan (-(90^\circ - \theta_2))\) [Using \(\tan (-A) = - \tan A\)]
\(\Rightarrow \tan \theta_1 = \tan (\theta_2 - 90^\circ)\)
\(\Rightarrow \theta_1 = \theta_2 - 90^\circ\) [since \(0^\circ \le \theta_1, \theta_2 \le 180^\circ\)]
\(\Rightarrow \theta_2 = \theta_1 + 90^\circ\) - - - - - (I)
An exterior angle of a triangle is equal to the sum of two opposite interior angles.
\(\Rightarrow \theta_2 = \angle A + \angle C\)
\(\Rightarrow \theta_2 = \theta_1 + \angle C\) - - - - - (II)
On comparing (I) and (II), we get:
\(\angle C = 90^\circ\)
Thus, the lines \(l_1\) and \(l_2\) are perpendicular.
Two lines are perpendicular if and only if the slopes are negative reciprocal of each other. That is, \(m_1 \cdot m_2 = -1\).
Example:
The slope of the line \(p\) is , and the slope of the line \(q\) is . Are the lines \(p\) and \(q\) are perpendicular?
Solution:
Let \(m_1\) be the slope of \(p\), and \(m_2\) be the slope of \(q\).
If \(m_1 \cdot m_2 = -1\), then the lines are perpendicular.
\(m_1 =\) \(=\) \(-4\)
\(m_2 =\) \(=\)
\(m_1 \cdot m_2 = - 4 \ \times \)
\(m_1 \cdot m_2 = - 1\)
Therefore, \(p\) and \(q\) are perpendicular lines.
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